Chapter 14, Problem 14.41P

(a)

Interpretation Introduction

Interpretation:

The condensed electron configurations for ${\text{In, In}}^{\text{+}}{\text{, In}}^{\text{2+}}$ and ${\text{In}}^{\text{3+}}$ has to be written.

Concept Introduction:

The condensed electron configurations:

The arrangement or filling of electrons in the orbitals of atom in known as electron configuration. The heaver atoms contains more number of orbitals and electrons so we can’t able to write full electron configuration so this time we can writhe the nearest noble gas then followed remaining electron configuration is known as condensed electron configuration.

Explanation of Solution

Atomic number of indium is $49$, and nearest noble gas is krypton therefore,

The condensed electron configurations of ${\text{In, In}}^{\text{+}}{\text{, In}}^{\text{2+}}$ and ${\text{In}}^{\text{3+}}$ are,

    $\begin{array}{l}\text{In}={\text{[Kr]4d}}^{\text{10}}{\text{,5s}}^{\text{2}}{\text{,5p}}^{\text{1}}\\ {\text{In}}^{+}={\text{[Kr]4d}}^{\text{10}}{\text{,5s}}^{\text{2}}\\ {\text{In}}^{2+}={\text{[Kr]4d}}^{\text{10}}{\text{,5s}}^{\text{1}}\\ {\text{In}}^{3+}={\text{[Kr]4d}}^{\text{10}}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The magnetism of ${\text{In, In}}^{\text{+}}{\text{, In}}^{\text{2+}}$ and ${\text{In}}^{\text{3+}}$ has to be given.

Concept Introduction:

Magnetism:

If an element contains unpaired of electrons, it is paramagnetic.

If an element does not contains unpaired of electrons, it is diamagnetic.

Explanation of Solution

The condensed electron configurations of ${\text{In, In}}^{\text{+}}{\text{, In}}^{\text{2+}}$ and ${\text{In}}^{\text{3+}}$ are,

    $\begin{array}{l}\text{In}={\text{[Kr]4d}}^{\text{10}}{\text{,5s}}^{\text{2}}{\text{,5p}}^{\text{1}}\\ {\text{In}}^{+}={\text{[Kr]4d}}^{\text{10}}{\text{,5s}}^{\text{2}}\\ {\text{In}}^{2+}={\text{[Kr]4d}}^{\text{10}}{\text{,5s}}^{\text{1}}\\ {\text{In}}^{3+}={\text{[Kr]4d}}^{\text{10}}\end{array}$

From the above electron configurations, ${\text{In and In}}^{\text{2+}}$ are contains unpaired of electros so they are paramagnetic.

${\text{In}}^{+}{\text{ and In}}^{\text{3+}}$ are doesn’t contains unpaired of electros so they are diamagnetic.

(c)

Interpretation Introduction

Interpretation:

The apparent oxidation state of $\text{In}$ in ${\text{InCl}}_{\text{2}}$ has to be given.

Concept Introduction:

Oxidation state:

The number of electrons lost or gain from an atom in the formation of molecule is known as oxidation number.

Explanation of Solution

In generally $\text{Cl}$ has $\text{-1}$ charge so the apparent oxidation state of $\text{In}$ in ${\text{InCl}}_{\text{2}}$ is,

    $\begin{array}{l}\text{x+2(-1)=0}\\ \text{x}\text{=}\text{2+}\\ \text{In}\text{=2+}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The diamagnetism of ${\text{InCl}}_{\text{2}}$ has to be explained.

Concept Introduction:

Oxidation state:

The number of electrons lost or gain from an atom in the formation of molecule is known as oxidation number.

Explanation of Solution

The condensed electron configurations of ${\text{In, In}}^{\text{+}}{\text{, In}}^{\text{2+}}$ and ${\text{In}}^{\text{3+}}$ are,

    $\begin{array}{l}\text{In}={\text{[Kr]4d}}^{\text{10}}{\text{,5s}}^{\text{2}}{\text{,5p}}^{\text{1}}\\ {\text{In}}^{+}={\text{[Kr]4d}}^{\text{10}}{\text{,5s}}^{\text{2}}\\ {\text{In}}^{2+}={\text{[Kr]4d}}^{\text{10}}{\text{,5s}}^{\text{1}}\\ {\text{In}}^{3+}={\text{[Kr]4d}}^{\text{10}}\end{array}$

From the above electron configurations, ${\text{In and In}}^{\text{2+}}$ are contains unpaired of electros so they are paramagnetic.

${\text{In}}^{+}{\text{ and In}}^{\text{3+}}$ are doesn’t contains unpaired of electros so they are diamagnetic.

In generally $\text{Cl}$ has $\text{-1}$ charge so the apparent oxidation state of $\text{In}$ in ${\text{InCl}}_{\text{2}}$ is,

    $\begin{array}{l}\text{x+2(-1)=0}\\ \text{x}\text{=}\text{2+}\\ \text{In}\text{=2+}\end{array}$

From the above discussion, ${\text{InCl}}_{\text{2}}$ contains half ${\text{In}}^{+}$ and half ${\text{In}}^{\text{3+}}$ ion therefore, ${\text{InCl}}_{\text{2}}$ is diamagnetic.